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Файл:The elements of the electrical circuit theory. Tutorial
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If the circuit contains m branches and n nodes, then the number of the
independent loops is
(1)
mn.
The circuit shown in Fig. 4.7 contains three independent loops.
5. THE EQUATIONS SET OF KIRCHHOFF’S LAWS
FOR CALCULATING THE CURRENTS OF THE CIRCUIT
Kirchhoff’s laws can be used for calculating currents in the circuit
branches. The main requirement in this case is to obtain the set of the independent equations in which the number of the unknown variables is equal to
the number of the currents to be calculated.
Nota bene! The following algorithm can be proposed for writing the
equations set of Kirchhoff’s laws:
1.
Count up the number of the nodes of the circuit. The number
of the equations of Kirchhoff’s current law is one less than the number
of the nodes. Choose the nodes for writing equations of Kirchhoff’s current law.
2.
Choose the independent loops. The easiest way is to choose them so
that they coincide with the cells of the circuit. Eliminate the loops with the
current sources. Choose the way around the loops.
3.
For the chosen loops, write the equations of Kirchhoff’s voltage law.
Add the equations of Kirchhoff’s current law to these.
4.
In the properly written the equations set, the number of the equations
is equal to the number of the unknown currents. The independence of the
equations is guaranteed by the independence of the loops and the fact that
the number of the equations of Kirchhoff’s current law is one less than the
number of the nodes.
Example 5.1 (writing the equations set of Kirchhoff’s laws)
The circuit for which it is necessary to write the equations set of Kirchhoff’s laws is shown in Fig. 5.1. The unknown currents are
1. There are four nodes in the circuit. Therefore, it is possible to write
three independent equations of Kirchhoff’s current law. Let us choose the
nodes
a, b and c.
21
II .
15
Fig. 5.1
2. There are three cells in the circuit. That is, the number of the independent loops is three. Let us choose the loops as Fig. 5.2 shows: the first
loop is
a-b-d-a; the second loop is b-c-d-b; the third loop is c-b-a plus the
current source.
Fig. 5.2
22
Let us eliminate the loop with the current source from the chosen loops.
R
R
R
There are the first loop and the second loop for the equation of Kirchhoff’s
voltage law. Let us choose the clockwise way around the loops.
3. Let us write the equations set:
I loop:
II loop:
a I I J
node: 0
b I I I
node: 0
c I I J
node: 0
IR I R I R E E
11 2 2 3 3 3 1
IR IR IR E
33 44 55 3
12
234
45
k
k
⎫
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎭
4. The number of the equations is equal to the number of the unknown
currents. Equations are independent. Therefore, the equations set has the
single solution.
In practice, it is often difficult to use the equations set of Kirchhoff’s
laws due to its large size. The more economical methods for calculating the
currents are developed based on Kirchhoff’s laws. At the same time, Kirchhoff’s laws are used to check the correctness of the solution and to find the
voltages between any points of the circuit, even if the branches do not connect these points to each other.
6. THE POWER BALANCE
OF THE ELECTRIC CIRCUIT
In electric circuit, the sources generate electric energy, and the energy
receivers consume it. In according to the law of conservation of energy, the
total energy of the sources should be equal to the total energy of consumers.
In the direct current circuit, all the energy (that is generated by the
sources) is consumed by the resistances (that convert electric energy to
heat). Joule’s law describes this conversation:
PIR (6.1)
cons
2
where
liberate out of itself per the unit of time);
with the consumer;
P is the consumer power (that is, the energy that the consumer
cons
I is the current in the branch
is the consumer resistance.
23
From the expression (6.1), it is obvious that the consumer power is always the positive value that does not depend on the direction of the current.
Let us now consider the sources of the electric energy. When the EMF
source gives the energy to the circuit, the current in the branch coincides
with the EMF source in the direction. In this case, the power is considered
positive (Fig. 6.1 a). The EMF source operates in the generator mode.
a) b)
Fig. 6.1
If the current in the branch is opposite to the EMF source in direction,
the EMF source does not give energy to the circuit, but consumes it. This
can happen, for example, when the accumulator gets charged. In this case,
the power is considered negative (Fig. 6.1 b). The EMF source operates in
the energy receiver mode.
In both cases, the following expression describes the power of the EMF
source:
where
P is the power of the EMF source; EI is the current in the branch
E
PEI (6.2)
EE
with the EMF source.
If the current coincides with the EMF source in the direction
(see Fig. 6.1 a), it is written in the formula (6.2) with the plus. If the current
is opposite to the EMF source in the direction (see Fig. 6.1 b), it is written in
the formula (6.2) with the minus.
The voltage at the terminals of the current source is used to calculating
the power of the current source. The positive direction of the voltage is the
direction from the terminal into which the current of the source enters to the
terminal from which the current of the source exits. In this case, the power
of the source is positive (Fig. 6.2 a).
24
J
J
J
R
a) b)
Fig. 6.2
If the voltage at the terminals of the current source is directed from the
terminal from which the current of the source exits to the terminal into
which the current of the source enters, the power of the source is negative
(Fig. 6.2 b).
In both cases, the following expression describes the power of the current source:
where
P is the power of the current source;
k
PJU (6.3)
kkJk
U is the voltage at the ter-
k
minals of the current source.
If the voltage at the terminals of the current source is directed in the
same direction as the current of the source (see Fig. 6.2 b), it is written in
the formula (6.3) with the minus. If the voltage at the terminals of the current source is opposite to the current of the source (see Fig. 6.2 a), it is written in the formula (6.3) with the plus.
The equality, which is called the power balance, is true for any electric
circuit:
PPP
∑∑∑
IR EI JU
∑∑∑
cons
2
EJk
EkJk
. (6.4)
Nota bene! It is necessary to note that in the formulas (6.4), to the left
of the equal sign, there is the arithmetic sum, and to the right of the equal
sign, there is the algebraic sum (that is, the sum in which the signs of the
terms are taken into account).
25
Nota bene! When we calculate the power of the sources in the balan-
ce (6.4), it is necessary, in the circuit diagram, to direct the arrows of the
currents and the voltages so that the supposed power is positive.
The power balance (6.4) is the independent check for all the analysis
methods of the electric circuits.
7. THE MESH-CURRENT METHOD
The mesh-current method is based on the assumption that the independent mesh-current flows in the independent loop of the electric circuit. It is
necessary to write equations for the mesh-currents, and the currents that do
exist in the branches of the circuit should be defined as the superposition of
the mesh-currents.
In the mesh-current method, the number of the unknown variables is
equal to the number of the equations that would need to be written for the
independent loops according to Kirchhoff’s voltage law. Therefore, the
mesh-current method is more economic than the method based on Kirchhoff’s laws (it has fewer equations).
Let us write the equations of the mesh-current method using the circuit
with two independent loops (Fig. 7.1).
Fig. 7.1
Let the mesh-current
current
I
flows in the right loop (also clockwise). First, let us write for-
22
I flow in the left loop clockwise, the mesh-
11
mulas that establish the relationship between the mesh-currents and the real
existing currents. There are two loops in the circuit. They have one common
26
branch (the serial connection of the resistance
R
E
Therefore, both the mesh-current of the first loop (
current of the second loop (
current
I flows from top to bottom, the mesh-current 22I flows from bot-
11
I ) flow in this branch. However, if the mesh-
22
and the EMF source
5
I ) and the mesh-
11
).
5
tom to top in this branch. The algebraic sum of the mesh-currents will be
equal to the real existing current in this branch, i.e. the current
into account the direction of the current
I
, we get the following:
2
I . Taking
2
II I. (7.1)
21122
Obviously, the real existing currents in the remaining branches (taking
into account the direction) are
II , (7.2)
111
II
322
. (7.3)
Let us write the equations of Kirchhoff’s voltage law for real existing
currents:
I loop: ( )
II loop: ( )
IR R IR E E
11 2 25 5 1
⎫
⎪
⎬
IR I R R E E
25 3 3 4 4 5
⎪
⎭
(7.4)
Now, let us substitute the expressions (7.1)-(7.3) into the equations set (8.4):
I loop: ( ) ( )
II loop: ( ) ( )
IRR I I R E E
11 1 2 11 22 5 5 1
IIRIRR EE
11 22 5 22 3 4 4 5
⎫
⎪
(7.5)
⎬
⎪
⎭
If we, in the equations set (7.5), regroup the terms, then we get the linear
equations set for the mesh-currents:
()
IRR R IR E E
11 1 2 5 22 5 5 1
IR I R R R E E
11522345 45
()
⎫
⎪
(7.6)
⎬
⎪
⎭
Nota bene! The following algorithm can be proposed for the analysis of
the electric circuit by means of the mesh-current method:
1.
Choose the arbitrary direction of mesh-currents in each of the inde-
pendent loops.
27
2. For each of the independent circuits, write the equations of Kirchhoff’s voltage law using the mesh-currents. In this case, Kirchhoff’s current
law is automatically true. In the properly written set of the equations, the
principal determinant is symmetric about the principal diagonal.
3.
After solving the equations set of the mesh-currents, define the real
existing currents in the branches as the algebraic sum of the mesh-currents.
If one mesh-current flows in the branch, then the real existing current is
equal to the mesh-current.
4.
The correctness of the solution can be verified with either the power
balance or Kirchhoff’s voltage law (but not Kirchhoff’s current law!).
Example 7.1 (using the mesh-current method)
In the circuit shown in Fig. 7.2, it is necessary to calculate the unknown
currents by means of the mesh-current method.
The given data are
5Ohm
R
1
10 Ohm
R
2
R
R
R
3
4
5
2Ohm
1Ohm
4Ohm
Fig. 7.2
R
R
E
E
E
28
6
7
2
4
5
5Ohm
2Ohm
10 V
8V
10 V
Let us choose the independent loops so that they coincide with the cells
R
R
R
R
R
R
R
R
R
R
R
R
R
of the circuit and let us direct all the mesh-currents clockwise. Let us write
the equations of Kirchhoff’s voltage law using the mesh-currents.
Let us consider the first loop. Its resistance consists of the resistors
and
6
opposite to the mesh-current
mesh-current
. In the branch with the resistor
5
I . This means that, for the first loop, the
22
I should be written with a minus in the equation of Kirch-
22
, the mesh-current
6
I
flows
11
1
hoff’s voltage law:
()IRR R IR E
11 1 6 5 22 6 5
. (Ex. 7.1.1)
Let us consider the second loop. Its resistance consists of the resistors
,
and
2
7
flows opposite to the mesh-current
with the resistor
current
I . This means that, for the second loop, both the mesh-current 11I
33
and the mesh-current
. In the branch with the resistor
6
I . At the same time, In the branch
11
, the mesh-current
7
I should be written with a minus in the equation of
33
I flows opposite to the mesh-
22
, the mesh-current 22I
6
Kirchhoff’s voltage law:
()IRRR IRIRE . (Ex. 7.1.2)
22 2 7 6 11 6 33 7 2
,
Let us consider the third loop. Its resistance consists of the resistors
and
4
opposite to the mesh-current
mesh-current
. In the branch with the resistor
7
I . This means that, for the third loop, the
22
I should be written with a minus in the equation of Kirch-
22
, the mesh-current 33I flows
7
,
3
hoff’s voltage law:
()IRR R IR E . (Ex. 7.1.3)
33 3 4 7 22 7 4
Let us join all the equations (Ex. 7.1.1)-(Ex. 7.1.3) into the equations
set. At the same time, let us pay attention to the fact whether all the unknown currents are in the equations and are located under each other:
()0
IRR R IR I E
⎧
⎪
⎪
⎨
⎪
⎪
⎩
11 1 6 5 22 6 33 5
IRIRRR IRE
11 6 22 2 7 6 33 7 2
0()
IIRIRRR E
11 22 7 33 3 4 7 4
()
29
(Ex. 7.1.4)
Let us substitute the numerical data into the equations set (Ex. 7.1.4):
14 5 0 10
II I
⎧
⎪
⎪
⎨
⎪
⎪
⎩
11 22 33
517 2 10
III
11 22 33
02 5 8
II I
11 22 33
(Ex. 7.1.5)
Nota bene! It is necessary to note that the principal determinant of is
symmetric about the principal diagonal in the equations set (Ex. 7.1.5).
Let us solve the equations set (Ex. 7.1.5) using Cramer’s method:
14 5 0
Δ 5 17 2 1009 Ohm
025
10 5 0
Δ 10 17 2 640 Volt×Ohm
1
825
14 10 0
Δ 510 2226Volt×Ohm
2
085
3
;
2
;
2
;
14 5 10
Δ 5 17 10 1524 Volt×Ohm
3
028
Δ
640
I
I
1
11
Δ 1009
Δ
226
22
Δ
2
Δ 1009
1524
3
Δ 1009
30
I
33
0.634 A
0.224 A
1.51 A
;
;
.
2
;
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